I was using Wolfram Alpha to compute this integral $$\frac{1}{\pi}\int_{-\pi}^{\pi}2\sin x \sin nx \; dx=\frac{4\sin\pi n}{\pi-\pi n^2}$$
So if $n\in\mathbb{Z}$ this answer must be $0$ right?
Then I used Wolfram Alpha again to compute $$\sum_{n=1}^\infty \frac{4\sin\pi n}{\pi-\pi n^2}\sin nx$$ and it says $$\sum_{n=1}^\infty \frac{4\sin\pi n}{\pi-\pi n^2}\sin nx=2\sin(x)$$ How is this possible?
Note that the first term in the sum is $$ \sin x \cdot \lim_{n\to1}\frac{4\sin \ \pi n}{\pi-\pi n^2} \\ \overset{\text{L.H.}}=4\sin x \cdot\ \lim_{n\to1} \frac{\cos \pi n\cdot \pi}{-2\pi n} \\ =2\sin x$$
The rest of the terms, as you say, will be zero.